Calculate the mass of compound (molar mass = `"256 g mol"^(-1)`) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (`K_(f)="5.12 K kg mol"^(-1)`)

Explanation

To solve the problem of calculating the mass of a compound to be dissolved in benzene to lower its freezing point, we will follow these steps:

### Step 1: Understand the formula for freezing point depression
The freezing point depression (\(\Delta T_f\)) is given by the formula:
\[ \Delta T_f = K_f \times m \]
where:
- \(\Delta T_f\) = depression in freezing point (in K)
- \(K_f\) = molal freezing point depression constant (in K kg/mol)
- \(m\) = molality of the solution (in mol/kg)

### Step 2: Calculate molality
Molality (\(m\)) is defined as:
\[ m = \frac{\text{mass of solute (in kg)}}{\text{mass of solvent (in kg)}} \]

### Step 3: Convert the mass of solvent from grams to kilograms
Given that the mass of benzene (solvent) is 75 g, we convert it to kg:
\[ \text{mass of solvent} = 75 \, \text{g} = 0.075 \, \text{kg} \]

### Step 4: Rearrange the freezing point depression formula to find molality
From the freezing point depression formula, we can rearrange it to find molality:
\[ m = \frac{\Delta T_f}{K_f} \]

### Step 5: Substitute the known values
Substituting the values into the equation:
\[ m = \frac{0.48 \, \text{K}}{5.12 \, \text{K kg/mol}} = 0.09375 \, \text{mol/kg} \]

### Step 6: Use molality to find the mass of solute
Using the definition of molality, we can express the mass of solute (\(m_{solute}\)) in terms of molality:
\[ m = \frac{m_{solute}}{M_{solute} \times \text{mass of solvent (in kg)}} \]
Rearranging gives:
\[ m_{solute} = m \times M_{solute} \times \text{mass of solvent (in kg)} \]

### Step 7: Substitute the known values
Substituting the values we have:
\[ m_{solute} = 0.09375 \, \text{mol/kg} \times 256 \, \text{g/mol} \times 0.075 \, \text{kg} \]

### Step 8: Calculate the mass of solute
Calculating this gives:
\[ m_{solute} = 0.09375 \times 256 \times 0.075 = 1.8 \, \text{g} \]

### Final Answer
The mass of the compound to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K is **1.8 g**.